Triangle P Q R is shown. The length of P Q is 17, the length of Q R is 15, and the length of P R is 14. Law of cosines: a2 = b2 + c2 – 2bccos(A) What is the measure of AngleP to the nearest whole degree? 35° 52° 57° 72°

P = 57°

Step-by-step explanation:

Given the following :

PQ = 17

QR = 15

PR = 14

Using the cosine formula since the length of the three sides are given:

a2 = b2 + c2 – 2bccos(A)

To find P:

QR^2 = PQ^2 + PR^2 – 2(PQ)(PR)cos(P)

15^2 = 17^2 + 14^2 – 2(17)(14)cos(P)

225 = 289 + 196 - 476 cosP

476*CosP = 485 - 225

476*CosP = 260

CosP = 260/476

CosP = 0.5462184

P = Cos^-1(0.5462184)

P = 56.892029

P = 57°

step-by-step explanation:

it's 15

given that triangle abc is transformed two times to get triangle

a"b"c"

first abc is transformed in to a'b'c' as follows:

the coordinates of a are (1,-1) and transformed into a'(-1,1)

similarly b (4,-2) became b'(-4,2) and

c(7,-2) became c'(-7,2)

i.e. (x,y) becomes (-x,-y)

this is nothing but reflection about a point here origin.

thus first transformation is reflection on the origin.

next is exactly shifting 3 units down

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